Figure shows as rod of length 20 cm pivoted near an end and which is made to rotate in a horizontal plane with a constant angular speed. A ball of mass m is spuspended by a string angular also of length 20 cm from the other end of the rod. If the angle `theta` made by the stirng with the vertical is `30^0`. find the anglular speed of the rotation. Take `g=10 m//s^2`

Options

A. `omega =5.0 rad//s`
B. `omega =9.8 rad//s`
C. `omega =2.2 rad//s`
D. `omega =4.4 rad//s`

Answer

D

Explanation

Let the be angular sped be `omega`. As ils clear from the ure the ball moves in a horizontal circlke of radius `L+L sin theta, where L=20 cm. `Its acceleration is therefore, `omega^2(L+Lsintheta)` thowareds the centre. The forces on the bob are ure.
a. The tension T along the string and
b. the weight mg. Resolving the forces along the radius and applying Newton's second law.
`Tsintheta=momega^2L(1+sintheta)` ............i
Applying Newton's first law in the vertical direction, `Tcostheta=mg` ,.........ii
Dividing i by ii
`tantheta=(omega^2L(1+sintheta))/g`
or ` omega^2=(g tantheta)/(L(1+sintheta))=((10 m/s^2)(1/sqrt3))/((0.20)(1+1/2))`
or, `omega =4.4 rad/s.